3.11.92 \(\int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx\) [1092]

Optimal. Leaf size=104 \[ \frac {b x^2}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^2}-\frac {\sqrt {b} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} \sqrt [4]{a+b x^4}} \]

[Out]

1/2*b*x^2/a/(b*x^4+a)^(1/4)-1/2*(b*x^4+a)^(3/4)/a/x^2-1/2*(1+b*x^4/a)^(1/4)*(cos(1/2*arctan(x^2*b^(1/2)/a^(1/2
)))^2)^(1/2)/cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x^2*b^(1/2)/a^(1/2))),2^(1/2))*b^(1
/2)/(b*x^4+a)^(1/4)/a^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {281, 331, 235, 233, 202} \begin {gather*} -\frac {\sqrt {b} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} \sqrt [4]{a+b x^4}}+\frac {b x^2}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^4)^(1/4)),x]

[Out]

(b*x^2)/(2*a*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(2*a*x^2) - (Sqrt[b]*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTa
n[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(2*Sqrt[a]*(a + b*x^4)^(1/4))

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 235

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + b*(x^2
/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{a+b x^2}} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^2}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^2}+\frac {\left (b \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx,x,x^2\right )}{4 a \sqrt [4]{a+b x^4}}\\ &=\frac {b x^2}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^2}-\frac {\left (b \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{4 a \sqrt [4]{a+b x^4}}\\ &=\frac {b x^2}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^2}-\frac {\sqrt {b} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 51, normalized size = 0.49 \begin {gather*} -\frac {\sqrt [4]{1+\frac {b x^4}{a}} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};-\frac {b x^4}{a}\right )}{2 x^2 \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^4)^(1/4)),x]

[Out]

-1/2*((1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, -((b*x^4)/a)])/(x^2*(a + b*x^4)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^4+a)^(1/4),x)

[Out]

int(1/x^3/(b*x^4+a)^(1/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(1/4)*x^3), x)

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Fricas [F]
time = 0.08, size = 25, normalized size = 0.24 \begin {gather*} {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{b x^{7} + a x^{3}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)/(b*x^7 + a*x^3), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.43, size = 31, normalized size = 0.30 \begin {gather*} - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**4+a)**(1/4),x)

[Out]

-hyper((-1/2, 1/4), (1/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(1/4)*x**2)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(1/4)*x^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\left (b\,x^4+a\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^4)^(1/4)),x)

[Out]

int(1/(x^3*(a + b*x^4)^(1/4)), x)

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